Simple Applications of Macroscopic
Thermodynamics

Preliminary Discussion
• Now, we drop the statistical mechanics notation for average quantities. So that now,
• We’ll discuss relationships between macroscopic variables using

• Some Thermodynamic Variables of Interest:
Internal Energy = E, Entropy = S
Temperature = T
(but also true for any substance):
External Parameter = V
Generalized Force = p
(V = volume, p = pressure)
External Parameter = x
Generalized Force = X

• Assume that the
External Parameter = Volume
V in order to have a specific case to discuss. For systems with another external parameter x , the infinitesimal work done đW = Xdx
. In this case, in what follows, replace p by X & dV by dx .
• For infinitesimal, quasi-static processes:
st
nd
1 st Law: đQ = dE + pdV
2 nd Law: đQ = TdS
st
nd

st
nd
• Note that, in this relation, there are
T, S, E, p, V
• It can be shown that:
Any 3 of these can always be expressed as functions of any 2 others.
• That is, there are always 2 independent variables & 3 dependent variables.
Which
2 are chosen as independent is arbitrary.

• Consider 3 variables: x, y, z.
Suppose we know that x & y are Independent
Variables .
Then, It Must Be Possible to express z as a function of x & y . That is,
.
• From calculus, the total differential of z(x,y) has the form:

y
x

• Suppose that, in this example of 3 variables: x, y, z, we want to take y & z as independent variables instead of x
& y . Then,
x = x(y,z) .
• From calculus, the total differential of x(y,z) is: dx

(∂x/∂y) z dy + (∂x/∂z)
• Using
(a) from the previous slide
[ dz

(∂z/∂x) y y dz (b) dx + (∂z/∂y) x dy (a) ]
& (b) together, the partial derivatives in (a) & those in
(b) can be related to each other.
• We always assume that all functions are analytic.
nd
Such as: (∂ 2 z/∂x∂y) 
(∂ 2 z/∂y∂x), etc.

Mathematics Summary
• Consider a function of 2 independent variables: f = f(x
1
,x
• It’s exact differential is df
 y
1 dx
1
2
).
+ y
2 dx
2
& by definition:
• Because f(x
1
,x that:
2
) is an analytic function, it is always true




 y x
1
2


 x
2




 y
 x
2
1


 x
1
•
Most Ch. 5 applications use this with the
Combined 1 st & 2 nd Laws of Thermodynamics
TdS = dE + pdV

Some Methods & Useful Math Tools for
Transforming Derivatives
Derivative Inversion



F
 y

 x


  y
1

F

 x

S

P

T


 
P
1

S


T

F
 x y


 x
 y


F
Triple Product (xyz–1 rule)
 y

F x
 
1

H

T
P

T

P
H

P

H

T
 
1
Chain Rule Expansion to Add Another Variable



F
 y

 x


F
  x


 
 y

 x

S

H
P


S

T
P

T

H
P

C
P
T
1
C
P

1
T


Maxwell Reciprocity Relationship
  
F
 y
 x
 y

 x

  
F
 x
 y
 x y
F xy

F yx

Pure Math: Jacobian Transformations
• A Jacobian Transformation is often used to transform from one set of independent variables to another.
• For functions of 2 variables
f(x,y) & g(x,y) it is:


 f x
,
, g
 


 f
 x
 g
 x y y





 f y
 g
 y

 x

 x

 f
 x y


 g
 y

 x



 f
 y

 x
 g
 x y

Jacobian Transformations
Have Several Useful Properties
Transposition
Inversion


 f x ,
, g
 

 


 g x ,
, f
 


 
 f x
,
, g y



 x
1 y



,
 f , g

Chain Rule
Expansion


 f x ,
, g
 



 
 f z
,
, g w

 

 z ,
, w
 


• Suppose that we are only interested in the first partial derivative of a function f(z,g) with respect to z at constant g:
 f
 z g



 f ,
, g
 

•
This expression can be simplified using the chain rule expansion & the inversion property


  f
 z


 g





 f x ,
, g
 

 
 

Properties of the Internal Energy E dE = TdS – pdV (1)
First, choose S & V as independent variables:
E

E(S,V)


U

S

V dS


U

V
S dV (2)
Comparison of (1) & (2) clearly shows that

S

V

T and

U

V
S
  p
Applying the general result with 2 nd cross derivatives gives:

T

V
S
 
 p

S

V

If S & p are chosen as independent variables , it is convenient to define the following energy:
H

H(S,p)

E + pV

Enthalpy
Use the combined 1 st & 2 nd Laws . Rewrite them in terms of dH: dE = TdS – pdV = TdS – [d(pV) – Vdp] or dH = TdS + Vdp
(1)
But, also:
Comparison of (1) & (2) clearly shows that and
(2 )
Applying the general result for the 2 nd cross derivatives gives:


 
T
 p 


S



 
V

S


 p

If T & V are chosen as independent variables , it is convenient to define the following energy:
F

F(T,V)

E - TS

Helmholtz Free Energy
•
Use the combined 1 st & 2 nd Laws . Rewrite them in terms of dF: dE = TdS – pdV = [d(TS) – SdT] – pdV or dF = -SdT – pdV
(1)
•
But, also: dF ≡ ( 
F/

T)
V dT + (

F/

V)
T dV (2)
•
Comparison of (1) & (2) clearly shows that
(

F/

T)
V
≡ -S and
(

F/

V)
T
≡ -p
•
Applying the general result for the 2 nd cross derivatives gives:

If T & p are chosen as independent variables , it is convenient to define the following energy:
G

G(T,p)

E –TS + pV

Gibbs Free Energy
•
Use the combined 1 st & 2 nd Laws . Rewrite them in terms of dH: dE = TdS – pdV = d(TS) - SdT – [d(pV) – Vdp] or dG = -SdT + Vdp
(1)
•
But, also: dG ≡ ( 
G/

T) p dT + (

G/
 p)
T dp
(2)
•
Comparison of (1) & (2) clearly shows that
(

G/

T) p
≡ -S and
(

G/
 p)
T
≡ V
•
Applying the general result for the 2 nd cross derivatives gives:

1. Internal Energy:
2. Enthalpy:
E

E(S,V)
H = H(S,p)

E + pV
3. Helmholtz Free Energy: F = F (T,V)

E – TS
4. Gibbs Free Energy: G = G(T,p)

E – TS + pV
nd
st
1. dE = TdS – pdV
2. dH = TdS + Vdp
3. dF = - SdT – pdV
4. dG = - SdT + Vdp

Another Summary:
Maxwell’s Relations
(a) ΔE = Q + W
(b) ΔS = (Q res
/T)
(c) H = E + pV
(d) F = E – TS
(e) G = H - TS

T

V
S
1.
 
 p

S

V

S

V

T
3.

 p

T

V



T
 p


S
2.


V

S



S
 p


T
4.
 

V

T p p
1. dE = TdS – pdV
2. dH = TdS + Vdp
3. dF = -SdT - pdV
4. dG = -SdT + Vdp
Mdx

Ndy
 dz



  z
 x 

 y dx



  z
 y 

 x dy



M
 y

 x



N x y

Maxwell Relations:
“The Magic Square”?
Each side is labeled with an
Energy (E, H, F, G) .
The corners are labeled with
Thermodynamic Variables
V
F
T
(p, V, T, S). Get the
Maxwell Relations by “walking” around the square. Partial derivatives are obtained from the sides.
E
S
The Maxwell Relations are obtained from the corners.
H
G
P



 
T

V 


S
 




P

S



V




S

V 


T



 
P

T



V


 
T

P 


S



 
V

S 


P



 
S

P



T



 
V

T 


P

Maxwell Relations: Table (E → U)

Maxwell Relations
Maxwell Relations from dE, dF, dH, & dG
Internal
Energy

T

V
S , N
 

P

S

V , N
Helmholtz
Free Energy

S

V

T , N


P

T

V , N
Enthalpy

T

P


V

S
S , N P , N
Gibbs Free
Energy

S

P

T , N
 

V

T
P , N

Some Common Measureable Properties
Heat Capacity at Constant Volume:
∂E
Heat Capacity at Constant Pressure:

More Common Measureable Properties
Volume Expansion Coefficient:
Note!!
Reif’s notation for this is α
Isothermal Compressibility:
The Bulk Modulus is the inverse of the Isothermal
Compressibility!
B

(κ) -1

Some Sometimes Useful Relationships
Summary of Results
Derivations are in the text and/or are left to the student!
Entropy:
Enthalpy:
Gibbs Free
Energy: d



G
RT 



V
RT dP

H
RT
2 dT

• Given the entropy
S as a function of temperature
T & volume V, S = S(T,V) , find a convenient expression for (

S/

T)
P
, in terms of some measureable properties.
• Start with the exact differential:
• Use the triple product rule & definitions:

• Use a
Maxwell Relation:
• Combining these expressions gives:
• Converting this result to a partial derivative gives:

• This can be rewritten as:
• The triple product rule is:
• Substituting gives:

•
Volume Expansion Coefficient

-1


p
•
Isothermal Compressibility

-1


T
•
Note again!!
Reif’s notation for the
Volume Expansion Coefficient is α

• Using these in the previous expression finally gives the desired result:
•
Using this result as a starting point,
A GENERAL RELATIONSHIP between the
Heat Capacity at Constant Volume C
V
& the
Heat Capacity at Constant Pressure C p can be found as follows:

•
Using the definitions of the isothermal compressibility κ and the volume expansion coefficient

, this becomes
General Relationship between C v
& C p

Simplest Possible Example: The Ideal Gas
• For an
Ideal Gas
, it’s easily shown
(Reif) that the
Equation of State
(relation between pressure P , volume V , temperature T ) is
(in per mole units!):
Pν = RT .
ν = (V/n)
• With this, it is simple to show that the volume expansion coefficient β & the isothermal compressibility κ are:
 
1 v


  v

T



P

1 v


 

T
RT
P



P

R vP

R
RT




1
T

1 v


  v

P



T and
 
1 v


 

P
RT
P



T

RT vP
2

RT
RTP
 
1
P

• So, for an
Ideal Gas , the volume expansion coefficient
& the isothermal compressibility have the simple forms: and
• We just found in general that the heat capacities at constant volume & at constant pressure are related as
• So, for an
Ideal Gas , the specific heats per mole have the very simple relationship:

Other, Sometimes Useful, Expressions
H
P .
T
S
P .
T
S
V .
T

P

P

0


V
 
P

P

0


 
P

P

0





T
V
T

P

T

V

V

T
P
P

 dP

R
P

 dP

R
V

 dV
CONSTANT T
CONSTANT T
CONSTANT T

More Applications: Using the Combined
1 st & 2 nd Laws
(“The TdS Equations”)
Calorimetry Again!
• Consider Two Identical Objects , each of mass m , & specific heat per kilogram c
P
. See figure next page.
Object 1 is at initial temperature T
1
.
Object 2 is at initial temperature T
2
.
Assume T
2
> T
1
.
• When placed in contact, by the 2 nd Law , heat Q flows from the hotter ( Object 2 ) to the cooler
( Object 1 ), until they come to a common temperature, T f
.

•
Two Identical Objects , of mass m , & specific heat per kilogram c
P
. Object 1 is at initial temperature T
1
.
Object 2 is at initial temperature T
2
.
• T
2
> T
1
. When placed in contact, by the 2 nd Law , heat Q flows from the hotter ( Object 2 ) to the cooler ( Object
1 ), until they come to a common temperature, T f
.
For some time Object 2
Q

Object 1 after initial Initially Heat Flows Initially contact: at T
2 at T
1
•
After a long enough time, the two objects are at the same temperature T f
. Since the 2 objects are identical, for this case,
T f

T
1

T
2
2

• The
Entropy Change
ΔS for this process can also be easily calculated:

S
 mc
P


 T
T
1 f dT
T
  T
T
2 f dT
T


 mc
P


 ln


T f
T
1


 ln


T f
T
2





 mc
P ln
T f
2
T
1
T
2
 mc
P ln



T f
T
1
T
2



2

2 mc
P ln
T f
T
1
T
2

S

2 mc
P ln
T
1

T
2
2 T
1
T
2
• Of course, by the 2 nd Law , the entropy change
ΔS must be positive!! This requires that the temperatures satisfy:
T
1

T
2

2 T
1
T
2
T
1
2 
T
2
2 
2 T
1
T
2
T
1
2 
T
2
2
( T
1

T
2
)
2

2 T
1
T
2

0

4 T
1
T
2

0

Some Useful
“TdS Equations”
•
NOTE: In the following, various quantities are written in per mole units! Work with the
Combined 1 st & 2 nd Laws:
Definitions:
•
•
• υ 
Number of moles of a substance.
• ν 
(V/υ)

Volume per mole.
• u

(U/υ)

Internal energy per mole.
• h

(H/υ)

Enthalpy per mole.
• s

(S/υ)

Entropy per mole. c c v
P

(C

(C v
P
/υ)
 const. volume specific heat per mole.
/υ)
 const. pressure specific heat per mole.

• Given these definitions, it can be shown that the Combined 1 st & 2 nd Laws (TdS) can be written in at least the following ways:
Tds
 c v dT

T
Tds
 c
P dT

T
Tds
 c
P

T
 v

P

T
 v

T v
P dv
 c v dv
 c v dT

T

 dv
P dP
 c
P dT

Tv
 dP

T

P v dP
 c

P v dv
 c v

 dP

• Student exercise to show that, starting with the previous expressions & using the definitions (per mole) of internal energy u & enthalpy h gives:
Internal Energy u(T,ν): du
Tds

 c
 u

T v dT v dT




 u
 v

T dv
 u
 v

T

P

 dv
Enthalpy h(T,P):

• Student exercise also to show that similar manipulations give at least the following different expressions for the molar entropy s : Entropy s(T,ν):
Consider s
 s ( P , v ) ds




 s
P s
P v v
 s

P

 dP
 c v
T v
 s

T v

T

P

 v
 s
 v
T
P v
P dv

1
T
T
 s

T v

T

P v ds ds
Tds


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P dv
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
T
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P v dP
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P

T
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P
P dv dv ds

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
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P
P
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P dP
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T
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T
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P
P

1
T
T
 s

T
P

T
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